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What volume of water at 0°C can a freezer make into ice cubes in 1.1 hours, if the coefficient of performa...?

What volume of water at 0°C can a freezer make into ice cubes in 1.1 hours, if the coefficient of performance of the Cooling Unit is 5.0 and the power input is 1.0 kilowatt?

Recall the definition of coefficient of performance, "what we want/what we pay for".

COP = Q_dot_in/W_dot_in
as in: rate of heat sucked in to the refrigerant piping/rate of electric work consumption

Heat which must be removed from ice:
Q = m*h_f, mass times heat of fusion

Since we were asked on a water volume basis:
Q = rho_water*V*h_f

And the rate at which the heat is sucked in to the refrigerant:
Q_dot_in = Q/t

Thus:
COP = (rho_water*V*h_f/t)/W_dot_in

Solve for V:
V = COP*W_dot_in*t/(rho_water*h_f)

Data
rho_water:=1000 kg/m^3; h_f:=334.9 kJ/kg; t:=1.1*3600 sec; COP:=5.0; W_dot_in:=1 kW;

Result:
V = 0.05912 m^3

Or can also be stated as:
V = 59.12 Liters

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